I What is the minimum force required to lift an object?

[Herman Trivilino]

When I reach the 100 N it will still remain because the net force I apply is zero So I need to go beyond the 100 N by any force greater than 100 N to lift the object.

When the object is moving upward at a steady speed the force you apply is 100 N. Not one bit more. A bit more would cause it to accelerate.

 

[Jodo]

I disagree. Let's say I applied a 50 N on the object it will not move and while I increase the force it will still remain. When I reach the 100 N it will still remain because the net force I apply is zero So I need to go beyond the 100 N by any force greater than 100 N to lift the object. The force is actually the smallest number greater than 100 N which is actually cannot be determined so when using the scale you cannot know whether the force is 100 N or greater because any force like 100.000001 N can cause non-zero net force and lift the object.

So a more accurate scale would prove this? One might wonder if 100.000001 N were needed to lift 100 N , would 100.00000997 work?
Something seems amiss here.

 

[Orodruin]

unless it is legday, then everything is allowed

What if every day is legday?

The force I need to throw the rock is hundreds times the force I need to jump even though the body and the rock are of the same mass.

No, it is not. Also keep in mind that it is not only a question of force but also of leverages. Your body is the result of millions of years of evolution and your legs shaped in such a way as to let you lift your body and move as efficiently as possible. It is less evolved to have the correct leverages to throw heavy objects. Lifting technique does help a lot but stones are also far less adapted to allow good lifting positions than say a barbell.

Back before Covid and becoming a father I used to powerlift for some time. I also helped my mother move some rocks at her summer house and I can tell you they seem much heavier for the same weight compared to a barbell simply because of worse leverages.

When your legs lift your body then that is all your body needs to do. If you also want to throw a heavy object then not only do your legs need to provide the force for both your weight and the object’s. The rest of your body, which as it has already been pointed out is significantly weaker, also needs to engage with typically bad leverage.

But if you lie on your back and can shove the rock with your legs, you might get a similar result to jumping (sending it maybe half a meter to a meter in the air).

Interestingly, there are gym machines designed specifically to do this in a more controlled manner:

I do not recommend loading it with so little that you can actually launch the foot plate in the air…

 

[malawi_glenn]

What if every day is legday?

Then you will never be able to throw that 60 kg stone and saying "ain't nothing put a peanut" whilst doing so

Interestingly, there are gym machines designed specifically to do this in a more controlled manner:

This is a quite more esotheric exercise, vertical legpress in a smith-machine

 

[Yahya Sharif]

This is a quite more esotheric exercise, vertical legpress in a smith-machine

This is an equivalent movement to jumping. The person is hardly lifting the load even though his muscles strength is above average. While he can jump without effort.

 

[malawi_glenn]

The person is hardly lifting the load even though his muscles strength is above average. While he can jump without effort.

What is your problem, really? What is that you do not understand about forces?
You want to know how much weight is put on that bar? Those were clean and easy reps, did not break a sweat!

You can also get muscle strength above average, just hit the gym bro instead of trying to deduce that the laws of physics does not apply to jumping or whatever you are trying to do here.

 

[Yahya Sharif]

When I try to lift my body as the experiment just before I lift my body the reading of the scale increases by the x force I mentioned in which the scale will read 60+x kgf when my body moves and accelerates upwards the x got a maximum of x2 N without accelerating upwards just my body at constant speed. What is this x? what its maximum value and why? The idea is for every body there is a minimum x force that can lift it. What is your opinions?

 

[russ_watters]

When I try to lift my body as the experiment just before I lift my body the reading of the scale increases by the x force I mentioned in which the scale will read 60+x kgf when my body moves and accelerates upwards the x got a maximum of x2 N without accelerating upwards just my body at constant speed. What is this x? what its maximum value and why? The idea is for every body there is a minimum x force that can lift it. What is your opinions?

I don't understand what you are asking. That's probably partly due to your long first sentence which reads like a run-on sentence and appears to contain multiple unconnected thoughts. And did you really mean to ask about minimum and maximum at the same time?

At this point though, you should understand this simple issue well enough to answer yourself - so why don't you tell us what you think the answers are? And maybe if you draw a diagram it will both help you understand the force balance and show us what you really mean by your scenario.

 

[Yahya Sharif]

In the video the weight is 1.29 kg. First I measured the weight then I started from zero Newtons to lift the weight the force continue to increase while I am trying to lift the weight. As soon as the weight raises the force reached its maximum 1.29 kgf in this case by definition the force I need to lift the 1.29 kg is 1.29 kgf which is equal to the weight.*

Now, in the case of the human body lifting his body, he starts pressing the scale, the x continues to increase, as soon as his body raises he reaches a maximum of x N then by definition the force to lift the body is the x N.

This x force is not a random force, it is a specific amount that is proportional to the human mass, If I repeat the lifting, as soon as my body raise I will get the same value if the human is a child of 25 kg the force in the scale to lift is smaller than x. So every human can lift his body with a specific force depending on his weight.

The x N force is the force that lifted the human body which is less than the weight 60 kg, in contrary of the physical fact that a force to lift a mass must be greater than the mass weight. This means a force smaller than the human body weight is sufficient to lift the body. What I mean is a human or an animal is an exception.
So what is this specific x N force ?*actually it is the smallest force greater than 1.29 kgf or 1.29+f the resultant is 1.29+f-1.29 or f, f is the force upwards I need this force to create a non-zero net force upwards. But as it is tiny it does not appear in the scale instead the scale will read 1.29 kg.

 

[jbriggs444]

The x N force is the force that lifted the human body which is less than the weight 60 kg, in contrary of the physical fact that a force to lift a mass must be greater than the mass weight.

Can you clarify, please.

You are saying that you exerted a force of less than 60 kg(force) and succeeded in lifting a human body with mass 60 kg(mass)? But you did not show us that experiment. Nor did you describe it.

 

[Yahya Sharif]

Can you clarify, please.

You are saying that you exerted a force of less than 60 kg(force) and succeeded in lifting a human body with mass 60 kg(mass)? But you did not show us that experiment. Nor did you describe it.

I didn't do the experiment it is my predictions. I don't have the big scale to measure my weight.
The description is in post #1

 

[russ_watters]

I didn't do the experiment it is my predictions.

Why would you predict something that contradicts the experiment you did? What is the reasoning behind it?

Just to be clear: the experiment and it's description/reasoning are fine. But then you appear to be discarding your own experiment and predicting the human body works differently. That makes no sense. And is pretty obviously wrong (unless you just didn't explain it correctly, which is possible...).

 

[russ_watters]

Back to your OP:

A person stands on a scale. The scale reads his mass 60 kg . Now this human moves up his body short distance like someone tries to pick a fruit from a tree. The scale will start to increase by small forces x N in which the total read of the scale is 600+x N *. The force he exerts on the scale is x N.

This is very clearly wrong. The scale reads the force that is exerted on it (600+xN). That's its purpose. Your own videoed experiment correctly demonstrates how it works. At this point I'm having trouble believing you are being serious.

 

[Yahya Sharif]

What is this specific fixed maximum value x N as soon as the human body raises ? What is the value of this x N for the 60 kg body and why? Why is the x force is smaller in case of a "child" of less weight of 25 kg?
In the video as soon as I lifted the weight the maximum force was 1.29 kgf which is the force to lift the mass. So what about a human lifting his body in the OP, what is this maximum x as soon as he lifts his body?

 

[jbriggs444]

What is this specific fixed maximum value x N as soon as the human body raises ?

Minimum, not maximum. Any positive, non-zero value of $x$ will result in the human body rising. Which, technically, means that there is no minimum. Only a greatest lower bound: zero.

 

[Yahya Sharif]

Minimum, not maximum. Any positive, non-zero value of $x$ will result in the human body rising. Which, technically, means that there is no minimum. Only a greatest lower bound: zero.

This is the scenario :
I press with x=10 N " don't move" I increase to x=15 N " still do not move" when I reach an x>15 N I raise. This is the maximum which is a fixed number if I repeat the experiment.
If any force can lift me why I cannot move with the x=10 N?

 

[jbriggs444]

This is the scenario :
I press with x=10 N " don't move" I increase to x=15 N " still do not move" when I reach an x>15 N I raise. This is the maximum which is a fixed number if I repeat the experiment.
If any force can lift me why I cannot move with the x=10 N?

Please specify the experiment. Since you mass more than 1.5 kg, it cannot be the one that you seem to be describing.

If I (hypothetically) mass 60 kg and weigh 600 N then a force (600+x) N in magnitude is sufficient to lift me when x = 1. Also when x = 0.1. Also when x = 0.01. Also when x = 0.001. Also when x = $10^{-100}$.

Pick a positive number. Any positive number. Then 600 N plus that number of Newtons will suffice to lift me. It does not take an excess of 10 N. It does not take an excess of 15 N. Any excess will do.

 

[russ_watters]

This is the scenario :
I press with x=10 N " don't move" I increase to x=15 N " still do not move" when I reach an x>15 N I raise. This is the maximum which is a fixed number if I repeat the experiment.
If any force can lift me why I cannot move with the x=10 N?

That isn't correct, unless you're changing the equation without telling us. It's F = 160 + x, where 160 is your weight. If that's still the equation, clearly what you are saying is wrong.

Again, you demonstrated how this works. Why do you think it works differently for people?

 

[Yahya Sharif]

If I (hypothetically) mass 60 kg and weigh 600 N then a force (600+x) N in magnitude is sufficient to lift me when x = 1. Also when x = 0.1. Also when x = 0.01. Also when x = 0.001. Also when x = $10^{-100}$.

Pick a positive number. Any positive number. Then 600 N plus that number of Newtons will suffice to lift me. It does not take an excess of 10 N. It does not take an excess of 15 N. Any excess will do.

But experiments do not say this. If you stand on the scale with weight 60 kg and you apply a smaller force let's say x=10 N the scale will read 600+10 N but you still do not rise. You will need to maximize your force to x N to lift your body.

 

[jbriggs444]

But experiments do not say this. If you stand on the scale with weight 60 kg and you apply a smaller force let's say x=10 N the scale will read 600+10 N but you still do not rise. You will need to maximize your force to x N to lift your body.

To be clear, you are standing on the scale and you push down on a nearby counter with a downward force of 10 N, you are concerned that you do not then rise up off of the scale?

Have you examined the scale in this circumstance to see whether the updated scale reading is now 590 N? The total supporting force would then be unchanged at... 600 N.

If you actually performed this experiment instead of thinking about it, you would discover that at x = 600 N, the scale reading would have gone to zero and you would then be lifting your feet free of the scale.

Edit: The actual behavior will depend crucially on whether the scale in question is a constant force device (one that will deliver a constant force of 600 N regardless of its vertical posision) or a constant position device (one that will deliver whatever force is required to maintain the position of the object resting upon it and will report that delivered force on a dial or digital readout).

A standard bathroom scale is, for most practical purposes, a constant position device.

Edit 2: Or, maybe this is what you are getting at...You are standing on a scale that reads 600 N. You kneel down and push with your hands, applying a downward force of 10 N with your hands on the top surface of the scale. You now expect the scale to read 610 N?

 

[russ_watters]

Edit 2: Or, maybe this is what you are getting at...

My interpretation of the scenario is jumping off the scale. The equation holds as originally described for that scenario. But the OP can clarify...

 

[Yahya Sharif]

Edit 2: Or, maybe this is what you are getting at...You are standing on a scale that reads 600 N. You kneel down and push with your hands, applying a downward force of 10 N with your hands on the top surface of the scale. You now expect the scale to read 610 N?

I do not kneel. I stand on the scale. my weight is 60 kg. I lift my body up short distance by feet and calves' muscles like someone trying to pick a fruit from a tree. If I press by a small force like 2 N I will not rise. The scale will read 600+2=602 N. I will need to maximize my force to some specific force x N to lift my body.
I start pressing the scale, the x continues to increase, as soon as my body raises I reach a maximum of x N then by definition the force to lift the body is the x N. This x has a specific value for the 60 kg If I repeat the experiment. If the person is " a child" of 25 kg the force x will be less.
In the video it is the same thing first I measured the weight which is 1.29 kg then I started to lift the weight, the force increases. As soon as the mass raises, the force reaches a maximum of 1.29 kgf that means by definition the force I used to lift the object equals the weight 1.29 kgf.
The force I used to lift the object in the video equals to its weight. The force I used to lift my body in the experiment x N is less than weight.
Humans and animals use force to lift or move their bodies smaller than the force they use to lift or move other objects with the same mass. So human movements such as : walking, running, jumping, dancing, etc, are done relatively with little effort.
So what is this maximum specific force x N that lifted the human?
P.S
The maximum force x is just a few Newtons compared to the weight 600 N. So the force to lift the 60 kg human body is just a few Newtons.

 

[Steve4Physics]

So what is this maximum specific force x N that lifted the human?

You mean minimum. And there is no limit.

There is some confusion here. I wonder if it's related to this...

If there is only a small change in force, internal friction in the scales may prevent the reading from changing. You will get a false reading.

But if you had perfect scales, then as soon as as you start to accelerate upwards (e.g. by straightening bent legs), the reading would increase.

If the acceleration is very small the increase will also be very small. E.g. a reading of 60.000kgf might increase to 60.001kgf.

 

[jbriggs444]

I do not kneel. I stand on the scale. my weight is 60 kg. I lift my body up short distance by feet and calves' muscles like someone trying to pick a fruit from a tree. If I press by a small force like 2 N I will not rise.

Wrong.

 

[Delta2]

Your CoM might rise without your feet leaving the ground. If your CoM is changing position then there is some net external force on your body.

 

[russ_watters]

I do not kneel. I stand on the scale. my weight is 60 kg. I lift my body up short distance by feet and calves' muscles like someone trying to pick a fruit from a tree. If I press by a small force like 2 N I will not rise. The scale will read 600+2=602 N. I will need to maximize my force to some specific force x N to lift my body.

That scenario is now clear, thanks for that at least. As others said, your understanding of the forces is wrong. Can you explain *why* you believe the physics works differently for humans than for the weight you tested?

Ultimately it's up to you if you want to learn/accept the reality or not. We can help, but there is only so much we can do if you won't explain your reasoning. This won't be productive if you just keep repeating the wrong thing over and over without explanation.

 

[berkeman]

Thread closed temporarily for Moderation...

 

[russ_watters]

Let's give the OP one more chance...

 

[Yahya Sharif]

The human will press the scale with x N. the scale will push with the same force x N then the force that lift the human is x N which is the force upwards by the scale.
A human legs undergoes a very small force upwards by the scale " x N" that results in very small pressure on the legs when the human stands on feet.
Observations:
1)a A human of 60 kg stands on his feet.
1)b The human is lying on the ground and legs are up and then putting a rock of 60 kg on the human's feet.
2)a The human is just lying on the ground.
2)b The human is lying on the ground on his belly a rock of 60 kg is put on his back, the rock has the human length and width.
In 1)a scenario the pressure on legs will be small that the human can stand for hours and this because the force upwards on the legs by the ground or a scale is a small force x N. In 1)b scenario the pressure will be extremely big and the legs cannot bare this 60 kg rock. This is also how a human knee can bare massive body 60 kg for years because the force on the legs is small x N.
In 2)a scenario the back will undergo small pressure that the human will be comfortable lying for hours. In 2)b scenario the rock will make very big pressure that it might break the chest. In all scenarios the human mass and the rock mass are equal 60 kg.

So the force that lifted the human In the experiment in the OP is a smaller force x N that appears as a smaller pressure on legs when standing or on the back when lying.

 

[russ_watters]

Okay, so your misunderstanding stems from the idea that because you do not have to exert much effort to stand up straight that means the force being applied to the ground is small. That is not the case. It just happens that our bodies are well configured for standing straight. In that condition we are essentially static objects. In other positions we are much less efficient at holding still. To see what I mean try this:

1. Start from a normal standing position.

2. Now bend your knees while keeping your torso vertical to lower your body a few centimeters until your knees are bent at an angle of 90°.

3. Hold this position until you understand the force that is actually required to be applied to the ground to hold you up. It should only take a few seconds.

 

[Ibix]

Another simple experiment you can do is to hold a weight of a few kilograms close to your chest, then extend your arms horizontally in front of you. I have a 10kg weight to hand. Close against my chest I can hold it for minutes with no trouble (I got bored before I got tired), but held out straight-armed in front of me I could only manage for a few seconds.

The force that needed to be exerted on the weight didn't change. The leverage my muscles had on my skeleton (and hence how hard they had to work) changed a lot.

Body mechanics is not simple introductory-level physics, and it's made complicated because you are only really aware of sensations out of the ordinary. For example, how do you think it would feel to have a 1kg weight pressing on 1cm² of your skin? Actually you already have that - atmospheric pressure - and you aren't aware of it. Another 1kg you would feel, and you would feel it differently if it were a single weight pressing on one place or if you doubled atmospheric pressure.

 

[Yahya Sharif]

Another simple experiment you can do is to hold a weight of a few kilograms close to your chest, then extend your arms horizontally in front of you. I have a 10kg weight to hand. Close against my chest I can hold it for minutes with no trouble (I got bored before I got tired), but held out straight-armed in front of me I could only manage for a few seconds.

The force that needed to be exerted on the weight didn't change. The leverage my muscles had on my skeleton (and hence how hard they had to work) changed a lot.

Body mechanics is not simple introductory-level physics, and it's made complicated because you are only really aware of sensations out of the ordinary. For example, how do you think it would feel to have a 1kg weight pressing on 1cm² of your skin? Actually you already have that - atmospheric pressure - and you aren't aware of it. Another 1kg you would feel, and you would feel it differently if it were a single weight pressing on one place or if you doubled atmospheric pressure.

Let's say for the the experiment in the OP the foot is 20 cm or 0.2 meters long , Now let's calculate for a 0.2 m lever:
First the lever will be class 2 :
The weight for 60 kg will be 60*10=600 Newtons.
Class 2 is the fulcrum at the toes , and in this case both the weight of my body and the force of my calves' muscles I lift my body with will be at the heel:
F: force of my weight
f: force of muscles strength
L: the distance of the weight from the heel to the toes.
l: distance of the muscles force from the heel to the toes.
f * l=FL
F=600 and L=l =0.2

f*0.2=600*0.2
f=600 Newton
The force needed to lift my body in the experiment does not change which is 600 N. I must exert a 600 N force to lift my body but I actually lift by weak calves' muscles and feet's muscles with the small force x N.

 

[hmmm27]

DRAW A DIAGRAM

of the simplest example you can think of that illustrates your problem with physics.

Label all forces.

 

[Yahya Sharif]

Why putting a rock of 60 kg on my back while I am lying on belly will give immediate extreme pressure that might break the chest but lying with human body mass 60 kg will cause small pressure that a human can lie for days ? The body and the rock are of the same weight. The mass of the rock has the same human shape so that the pressure by the rock on the back will be equivalent to the pressure by the human body on the back.

 

[jbriggs444]

Let's say for the the experiment in the OP the foot is 20 cm or 0.2 meters long , Now let's calculate for a 0.2 m lever:

So it seems that you will be working to calculate the force of the calf muscle as concentrated in the Achilles tendon.

First the lever will be class 2 :
The weight for 60 kg will be 60*10=600 Newtons.

Right. 600 Newtons upward force from the ground on the foot centered near the toes. Hypothetically 20 cm forward from the ankle joint.

Class 2 is the fulcrum at the toes , and in this case both the weight of my body and the force of my calves' muscles I lift my body with will be at the heel:

Right so you envision the toes as the fulcrum for this lever. Not the best spot to use to illustrate the principle, but let us see where you are going with this.

F: force of my weight
f: force of muscles strength
L: the distance of the weight from the heel to the toes.
l: distance of the muscles force from the heel to the toes.

Ok. Achilles tendon attaches to foot at distance lower case "l" from the toes.
Meanwhile, the weight of the body attaches to the foot at a distance upper case "L" from the toes.

f * l=FL
F=600 and L=l =0.2

Wait a minute. You are saying that the attachment point for the Achilles tendon and the center of the ankle joint are co-located. Please consult a drawing of a real foot.

Personally, I would have treated this as a lever with the fulcrum at the ankle. But then I'd have first realized that there is no need to treat the foot as a lever at all. We have a body with a center of mass and we have an external force. We don't need any levers to describe the resulting motion.

 

[Ibix]

Why putting a rock of 60 kg on my back while I am lying on belly will give immediate extreme pressure that might break the chest but lying on back with human body mass 60 kg will cause small pressure that a human can lie for days even though the body and the rock are of the same mass?

I think you are over estimating how much pressure a 60kg rock 2m long and say 50cm wide will exert. It's 600Pa, which is about the same as water pressure at a depth of 6cm - easily survivable. The rock would be between one and two centimeters thick, depending on what rock you have in mind.

Of course the top surface of a supine human is not flat and some points will suffer a higher pressure, so I'd expect it to be uncomfortable (rather like lying face down on flat concrete is, in fact) but hardly fatal.

The force needed to lift my body in the experiment does not change which is 600 N. I must exert a 600 N force to lift my body but I actually lift by weak calves' muscles and feet's muscles with the small force x N.

You are missing the point of both @russ_watters' comment and mine. The point is that your sensation of how hard or painful it is to do something is a terrible guide to how much force is needed. Depending on how you arrange your skeleton, the amount of force you need to apply to hold a fixed mass stationary can vary wildly. That isn't because humans are an exception to the laws of physics, it's because we can change our configuration so that the force we need to remain in a given configuration varies.

 

[Yahya Sharif]

View attachment 305068

Personally, I would have treated this as a lever with the fulcrum at the ankle. But then I'd have first realized that there is no need to treat the foot as a lever at all. We have a body with a center of mass and we have an external force. We don't need any levers to describe the resulting motion.

I am not a native speaker I really meant by the heel the joint of the ankle. I can make my body perfectly vertical when I balance my body holding something then the idea of a lever will work. The force must lift the human is still 600 N.

 

[Yahya Sharif]

I think you are over estimating how much pressure a 60kg rock 2m long and say 50cm wide will exert. It's 600Pa, which is about the same as water pressure at a depth of 6cm - easily survivable.

Of course the top surface of a supine human is not flat and some points will suffer a higher pressure, so I'd expect it to be uncomfortable (rather like lying face down on flat concrete is, in fact) but hardly fatal.

There is a difference between the pressure under water and the pressure of a mass on body.
At 10 meters under water the pressure on the body is 1 atmosphere which is equivalent to 101325 pascal the human will survive.

https://www.google.com/search?q=atm....69i57j0l5.12585j0j7&sourceid=chrome&ie=UTF-8

But 101325 pascal or N/m^2 is a rock of 10132.5 kg on the human body of 2 m length and 50 cm width so a mass of 10 tonnes is on the human body will crush all human body bones.

You are missing the point of both @russ_watters' comment and mine. The point is that your sensation of how hard or painful it is to do something is a terrible guide to how much force is needed.

Damage. Damage measures the severity of pressure on the human. Compare putting a protrusion of 18 cm width 18 cm length of a rock of 60 kg on belly vs lying on a concrete protrusion of 18 cm width 18 cm length on belly and my rest of the body is on air.
In case of human you will feel some pain but in the case of a rock the rock will damage the belly completely.

 

[malawi_glenn]

But 101325 N/m^2 is a rock of 10132.5 kg on the human body of 2 m length and 50 cm width so a mass of 10 tonnes is on the human body will crush all human body bones.

You can not make that rock have 1 m² contact on the human so the pressure will be much larger.

Humans can be buried in sand which are tiny tiny rocks. But the weight of the sand is more evenly distributed along the human body thus increasing the area of contact.

Damage measures the severity of pressure on the human. Compare putting a protrusion of 18 cm width 18 cm length of a rock of 60 kg on belly vs lying on a concrete protrusion of 18 cm width 18 cm length on belly and my rest of the body is on air.
In case of human you will feel some pain but in the case of a rock the rock will damage the belly completely.

Eh, no it won't crush your belly...

 

[Ibix]

do not know

And this is the problem. You are guessing instead of calculating. Of course, you cannot calculate because you have no model except your own personal incredulity that human bodies obey the laws of physics.

there is a different between the pressure under water and the pressure of a mass on body.

There is. It's that water pressure is isotropic (more or less) and the rock isn't. Nevertheless, it's worth comparing your rock to water. If I put a child's paddling pool on you and filled it to a depth of 6cm, do you think it would hurt? If I put a 60kg 2m×0.5m rock on you do you think it would hurt? You've already answered yes to the second question, and you now know the paddling pool weighs the same. But what's your instinctive response to the paddling pool? I bet it's not "that would destroy me" - but if your instincts are a consistent physical model then they should tell you the paddling pool with 6cm of water will crush you the same as the rock.

Of course, neither will crush you.

At 10 meters under water the pressure on the body is 1 atmosphere which equivalent to 101325 pascal

And I expect a "paddling pool" ten meters high filled with water and placed on your chest would crush you, yes.

Compare putting a protrusion of 18 cm with 18 cm length of a rock of 60 kg on belly vs lying on a concrete protrusion of 18 cm width 18 cm length on belly and my rest of the body is on air.
In case of human you will feel some pain but in the case of a rock the rock will damage the belly completely.

Nonsense. Go to YouTube and search for "weightlifting fails". The site is awash with videos of idiots benchpressing too much without a spotter and ending up trapped under their barbells' bars. That easily provides pressures well above your 18×18 rock example, even if you neglect the initial extra pressure from stopping the weights' fall. It doesn't look comfortable, and I bet they have impressive bruising, but they're not cut in half.

 

[malawi_glenn]

Go to YouTube and search for "weightlifting fails"

I hope my videos do not show up, would be embarrassing

If I put a 60kg 2m×0.5m rock on you do you think it would hurt?

The spatial size of the rock is irrelavant, what matters is the area of contact. It will be pretty hard to find a rock with 60 kg mass which has a flat side with dimensions 2m x 0.5m

Of course the top surface of a supine human is not flat and some points will suffer a higher pressure, so I'd expect it to be uncomfortable (rather like lying face down on flat concrete is, in fact) but hardly fatal.

 

[Orodruin]

videos of idiots benchpressing too much without a spotter and ending up trapped under their barbells' bars

Don’t look at the squat fails if you don’t want to scream in empathy pain…

The bottom line though is that the OP is trying to equate experienced sensation to physical measurement. This is of course bound to fail because sensation depends on so much more - as already pointed out repeatedly. The OP won’t learn anything by stubbornly refusing to accept the shortcomings of their approach.

A very small force is enough to kill a human if applied appropriately - like at the tip of a knife. The same force spread evenly over the body, not so much.

 

[Ibix]

The spatial size of the rock is irrelavant, what matters is the area of contact. It will be pretty hard to find a rock with 60 kg mass which has a flat side with dimensions 2m x 0.5m

Well, you could get one cut. As I noted upthread it would be only a centimeter or two thick, so actually pretty fragile and would need careful handling.

I suspect that's why the OP has now switched to talking about 18cm × 18cm rocks (about the same size as a pair of human feet): he wants to compare to someone standing on your stomach. Neither experience would be particularly comfortable, I think, but neither would crush you to a pulp.

I hope my videos do not show up, would be embarrassing

I'm going to google "malawi_glenn weightlifting fails" now...

Don’t look at the squat fails if you don’t want to scream in empathy pain…

I'll bear that in mind...

A very small force is enough to kill a human if applied appropriately - like at the tip of a knife.

The button on the tip of a sport fencing sword should trigger if held vertically with a 200g weight on top (you can get cylindrical weights with a coaxial hole drilled for testing). According to official fencing rules that's all the force you need to stab someone with a pointed weapon.

 

[Orodruin]

According to official fencing rules that's all the force

Do the official rules specify standard gravity?

 

[Ibix]

Do the official rules specify standard gravity?

I don't know, actually. As long as everyone's swords are tested at the same altitude for a given competition nobody has an advantage, though.

Actually, I think I have it wrong. The button shouldn't trigger with a 200g weight, but may trigger with a 200+$\epsilon$ gram weight.

 

[Orodruin]

I'm going to google "malawi_glenn weightlifting fails" now...
I'll bear that in mind...

Things not to search for on youtube:
”Squat fail knee”
”Deadlift fail biceps” (what? Not supposed to try to biceps curl 300+ kg??)

 

[Orodruin]

I hope my videos do not show up, would be embarrassing

So you do have fail videos!

The big question of course becomes: What is your powerlift total and how much did you put on the bar when you failed?

 

[Yahya Sharif]

The bottom line though is that the OP is trying to equate experienced sensation to physical measurement.

In post #88 I used the idea of measuring by damage to the body not sensation. Damage to the body can be a physical measurement. It is whether the rock damage the belly or not.
The rock of 60 kg on belly area 18 cm length and 18 cm width will extremely damage the belly. A human lying on belly on a protrusion on the ground of 18 cm length and 18 cm width will only cause some pain"no damage"

A very small force is enough to kill a human if applied appropriately - like at the tip of a knife. The same force spread evenly over the body, not so much.

Yes, a knife with small force will kill a human.This is because the tip of the knife is of very small area. The area of the rock on the belly is exactly the belly dimensions 18 cm length and 18 cm width which is bigger. So the rock will give smaller damage than the knife because the area of the rock is bigger.

Eh, no it won't crush your belly...

Can you dare and put a rock of 60 kg on your belly?

 

[malawi_glenn]

Can you dare and put a 60 kg rock on your belly?

Yes.

Why do you think it is something special about rocks?


nice hollywood special effects, right?

 

[Orodruin]

In post #88 I used the idea of measuring by damage to the body not sensation.

Which, as has also been pointed out, is a very poor alternative because that is even more dependent on how the force is applied. You are only digging your hole deeper and deeper.

The rock of 60 kg on belly area 18 cm length and 18 cm width will extremely damage the belly.

No, it will not. Stop making things up.

The heaviest deadlift on record is 501 kg. By your argumentation this should have ripped the guy’s arms off.

Yes, a knife with small force will kill a human.This is because the tip of the knife is of very small area. The area of the rock on the belly is exactly the belly dimensions 18 cm length and 18 cm width which is bigger. So the rock will give smaller damage than the knife because the area of the rock is bigger.

Exactly. So you first say to use damage as a measurement and in the next sentence agree that damage is highly dependent on mode of application. Now you are just being inconsistent. That won’t do.